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3 years ago

A coffee shop s ends $408 for an order of 17 cases of paper cups.

Mathematics

1 answer:

timama [110]3 years ago

7 0

That would be 528 my friend

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HELP ASAP!!! ILL GIVE LOTS OF POINTS AND BRAINLIST<br><br> PLEASE SHOW HOW YOU GOT THE ANSWER!

Kruka [31]

Answer: 27

Step-by-step explanation:

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2 years ago

I need help with question 3 and 4

katrin2010 [14]

Question 3. You divide 5 1/4 by 3/4 then multiply that by 20.
The answer is 140

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4 years ago

Lizette is training for a marathon. At 7:00 she left her house and ran until 8:30, then she walked until 11:30. She covered a to

Leokris [45]

<h3>Answer:</h3>

running speed: 8 mph
walking speed: 2 mph

<h3>Step-by-step explanation:</h3>

Let w represent Lizette's walking speed. Then her running speed is w+6. The relationship between speed, time, and distance is ...

distance = speed × time

Lizette ran for 1.5 hours, then walked for 3 hours. Her total distance is ...

(w+6)·1.5 + w·3 = 18

4.5w + 9 = 18 . . . . . simplify

4.5w = 9 . . . . . . . . . subtract 9

w = 2 . . . . . . . . . . . .divide by 4.5

Lizette's walking speed is 2 mph; her running speed is 8 mph.

6 0

3 years ago

Read 2 more answers

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

A shop buys a pair of shoes for $50. They sell them for $60. What is the mark–up on the shoes?

mojhsa [17]

Answer:

The markup for the shoes is 20 percent.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers